Answer the following questions for the function

f(x) = sin^2(x/3)
defined on the interval [ -9.424778, 2.356194].

Rememer that you can enter pi for \pi as part of your answer.

a.) f(x) is concave down on the interval .

b.) A global minimum for this function occurs at .

c.) A local maximum for this function which is not a global
maximum occurs at .

d.) The function is increasing on the region .

We went through this already:

Posted by Anonymous on Tuesday, November 1, 2011 at 1:53pm.

f(x) = sin^2(x/2)
defined on the interval [ -5.683185, 1.270796].

Take a look at that one for the method.

i cant see how you figure that out

I cant find the interval in which it concaves down, and that's the only question i need.

f = sin^2(x/3)

f' = 2/3 sin(x/3)cos(x/3) = 1/3 sin(2x/3)
f'' = 2/9 cos(2x/3)

f is concave down when f'' < 0

so, where is cos(2x/3) < 0?
When pi/2 < 2x/3 < 3pi/2
or 3pi/4 < x < 9pi/4

Your interval appears to be about (-3pi/2,pi/2), so you may have to adjust your answer a bit to fit the interval.

To determine the answers to the questions, we need to analyze the given function f(x) = sin^2(x/3) on the interval [ -9.424778, 2.356194].

a) To determine if f(x) is concave down, we need to check the second derivative of the function. If the second derivative is negative, then the function is concave down.

First, let's find the first derivative of f(x):
f'(x) = 2*sin(x/3) * (1/3) * cos(x/3)

Now, let's find the second derivative:
f''(x) = [d/dx(2*sin(x/3) * (1/3) * cos(x/3))] = (2/3) * cos^2(x/3) - (2/3) * sin^2(x/3)

To check if f(x) is concave down, we need to evaluate f''(x) at any point within the given interval [-9.424778, 2.356194].

For simplicity, let's evaluate f''(0):
f''(0) = (2/3) * cos^2(0/3) - (2/3) * sin^2(0/3)
= (2/3) * (1) - (2/3) * (0)
= 2/3

Since f''(0) = 2/3 > 0, f(x) is not concave down on the given interval.

b) To find the global minimum, we need to find the critical points of f(x) within the interval [-9.424778, 2.356194]. Critical points occur where the first derivative is equal to zero or undefined.

Setting f'(x) = 0:
2*sin(x/3) * (1/3) * cos(x/3) = 0

This equation holds true when sin(x/3) = 0 or cos(x/3) = 0.

sin(x/3) = 0 when x/3 = 0, pi, 2*pi, ...

cos(x/3) = 0 when x/3 = pi/2, 3*pi/2, 5*pi/2, ...

Within the interval [-9.424778, 2.356194], none of these critical points fall within the interval, so we cannot determine the global minimum.

c) To find a local maximum that is not a global maximum, we can analyze the critical points of f(x) within the interval [-9.424778, 2.356194]. We already found the critical points in part b.

From the critical points obtained, we can ignore the values that fall outside the interval [-9.424778, 2.356194].

Within the given interval, the critical points are not local maxima, so there is no local maximum that is not a global maximum.

d) To determine where the function is increasing, we need to analyze the sign of the first derivative.

From part a, we already found the first derivative of f(x):
f'(x) = 2*sin(x/3) * (1/3) * cos(x/3)

To determine where the function is increasing, we need to find the intervals where f'(x) > 0. This indicates that the function is positive and, therefore, increasing in those intervals.

Analyzing the sign of f'(x) throughout the given interval [-9.424778, 2.356194] is best done visually or with a graphing tool. However, let's take a closer look at the critical points we found earlier.

From part b, we found the critical points:
sin(x/3) = 0 when x/3 = 0, pi, 2*pi, ...
cos(x/3) = 0 when x/3 = pi/2, 3*pi/2, 5*pi/2, ...

Within the given interval, the function f(x) crosses the x-axis at x = pi, 2*pi, and 3*pi.

Therefore, the function is increasing in the intervals: (-9.424778, pi), (2*pi, 3*pi), and (3*pi, 2.356194).

Please note that this analysis is based on the information provided, and it is important to double-check using appropriate tools or graphs for more accurate results.